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## 'Square Mean' printed from http://nrich.maths.org/

Well done Freddie from Packwood Haugh
School and Danny from Milliken High School, Canada.

Is the mean of the squares of two numbers greater than, or less
than, the square of their means? Let the two numbers be $ p$ and $
q $.

$$\text{Square of mean } = {( p + q )^2\over 4} = {{p^2 + 2pq +
q^2}\over 4} $$

$$\text{Mean of squares } = {{p^2 + q^2}\over 2}
$$.

$$ {{p^2 + q^2}\over 2} - {{p^2 + 2pq + q^2}\over 4} = {{p^2 -2pq +
q^2}\over 4}= {(p - q)^2\over 4}\geq 0. $$.

Note that this difference, $ {(p - q)^2\over 4} $, is zero if $ p=q
$ and positive for all other choices of $ p $ and $ q $. So if the
numbers are equal then the mean of the squares is equal to the
square of the mean. Otherwise the mean of the squares is greater
than the square of the mean.